在SQLAlchemy中查询混合属性

10 投票
1 回答
4911 浏览
提问于 2025-04-17 13:32

我在数据库里存储的是相对路径的文件路径,但我使用混合属性把它转换成绝对路径。当我用这个属性查询时,它会报错。下面是我的模型:

class File(Base):
    __tablename__ = 'files'
    ...

    _f_path = Column(Unicode(30))

    ...

    @hybrid_property
    def f_path(self):
        env = shelve.open('environment')
        return os.path.join(env['project_dir'], self._f_path)

    @f_path.setter
    def f_path(self, _f_path):
        self._f_path = _f_path

当我运行这个查询(其中 ref 是一个 Unicode 字符串):

session.query(File).filter_by(f_path=ref).first()

它给我报了这个错:

File "/Users/Ben/Dropbox/Giraffe/giraffe_server/giraffe/file_handlers/maya.py", line 135, in process_file
    rf = session.query(File).filter_by(f_path=str(ref)).first()
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/orm/query.py", line 1211, in filter_by
    for key, value in kwargs.iteritems()]
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/orm/util.py", line 597, in _entity_descriptor
    return getattr(entity, key)
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/ext/hybrid.py", line 681, in __get__
    return self.expr(owner)
  File "/Users/Ben/Dropbox/Giraffe/giraffe_server/giraffe/model.py", line 133, in f_path
    print "\n\n\n[model.py:File@f_path hybrid_property] returning: ", os.path.join(env['project_dir'], self._f_path)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/posixpath.py", line 66, in join
    if b.startswith('/'):
  File "build/bdist.macosx-10.7-intel/egg/sqlalchemy/sql/expression.py", line 3426, in __nonzero__
    raise TypeError("Boolean value of this clause is not defined")
TypeError: Boolean value of this clause is not defined

1 个回答

15

你的混合属性应该返回一个SQL表达式,但你现在返回的是一个Python字符串。

要解决这个问题,不要在Python中进行路径连接,而是应该在SQL表达式中进行:

return env['project_dir'] + os.path.sep + self._f_path

这样就会变成 self._f_path.__radd__(result_of_project_dir_plus_os_path_sep),这个结果可以在查询中使用,也可以作为返回值。

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